Difference between revisions of "Gossard perspector"
(→Euler line of the triangle formed by the Euler line and the sides of a given triangle) |
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Denote <math>\angle ABC = \beta, \angle ACB = \gamma.</math> | Denote <math>\angle ABC = \beta, \angle ACB = \gamma.</math> | ||
− | It is known that < | + | It is known that <cmath>\tan \angle AED = \tan \theta_B = \frac{3 – \tan \alpha \cdot \tan \gamma}{\tan \alpha – \tan \gamma}.</cmath> |
− | <cmath>\tan \ | + | *[[Euler line]] |
+ | <cmath>\tan \angle AED = \frac{3 – \sqrt{3} \tan \gamma}{\sqrt{3} – \tan \gamma} = \sqrt{3} \implies \angle AED = 60^\circ.</cmath> | ||
Therefore <math>\triangle ADE</math> is equilateral triangle. | Therefore <math>\triangle ADE</math> is equilateral triangle. | ||
− | + | ||
− | Let <math>\triangle A'B'C'</math> be the triangle formed by the Euler lines of the <math>\triangle BDF, \triangle CEF,</math> and the line <math>l</math> contains centroid <math>G</math> of the <math>\triangle ADE</math> and parallel to <math>BC,</math> the vertex <math>B'</math> being the intersection of the Euler line of the <math>\triangle CEF</math> and <math>l,</math> the vertex <math>C'</math> being the intersection of the Euler line of the <math>\triangle BDF</math> and <math>l,</math> the vertex <math>A'</math> being the intersection of the Euler lines of the <math>\triangle BDF</math> and <math>\triangle CEF.</math> | + | Let <math>\triangle A'B'C'</math> be the triangle formed by the Euler lines of the <math>\triangle BDF, \triangle CEF,</math> and the line <math>l</math> contains centroid <math>G</math> of the <math>\triangle ADE</math> and parallel to <math>BC,</math> the vertex <math>B'</math> being the intersection of the Euler line of the <math>\triangle CEF</math> and <math>l,</math> the vertex <math>C'</math> being the intersection of the Euler line of the <math>\triangle BDF</math> and <math>l,</math> the vertex <math>A'</math> being the intersection of the Euler lines of the <math>\triangle BDF</math> and <math>\triangle CEF.</math> |
+ | |||
We call the triangle <math>\triangle A'B'C'</math> as the Gossard triangle of the <math>\triangle ABC.</math> | We call the triangle <math>\triangle A'B'C'</math> as the Gossard triangle of the <math>\triangle ABC.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 05:57, 15 January 2023
Contents
Gossard perspector X(402) and Gossard triangle
In Leonhard Euler proved that in any triangle, the orthocenter, circumcenter and centroid are collinear. We name this line the Euler line. Soon he proved that the Euler line of a given triangle together with two of its sides forms a triangle whose Euler line is in parallel with the third side of the given triangle.
Professor Harry Clinton Gossard in proved that three Euler lines of the triangles formed by the Euler line and the sides, taken by two, of a given triangle, form a triangle which is perspective with the given triangle and has the same Euler line. The center of the perspective now is known as the Gossard perspector or the Kimberling point
Let triangle be given. The Euler line crosses lines
and
at points
and
On it was found that the Gossard perspector is the centroid of the points
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Gossard perspector of right triangle
It is clear that the Euler line of right triangle meet the sidelines
and
of
at
and
where
is the midpoint of
Let be the triangle formed by the Euler lines of the
and the line
contains
and parallel to
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler line of the
and
We call the triangle as the Gossard triangle of
Let be any right triangle and let
be its Gossard triangle. Then the lines
and
are concurrent. We call the point of concurrence
as the Gossard perspector of
is the midpoint of
is orthocenter of
is circumcenter of
so
is midpoint of
is the midpoint
is the midpoint
with coefficient
Any right triangle and its Gossard triangle are congruent.
Any right triangle and its Gossard triangle have the same Euler line.
The Gossard triangle of the right is the reflection of
in the Gossard perspector.
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Gossard perspector and Gossard triangle for isosceles triangle
It is clear that the Euler line of isosceles meet the sidelines
and
of
at
and
where
is the midpoint of
Let be the triangle formed by the Euler lines of the
and the line
contains
and parallel to
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler line of the
and
We call the triangle as the Gossard triangle of
Let be any isosceles triangle and let
be its Gossard triangle. Then the lines
and
are concurrent. We call the point of concurrence
as the Gossard perspector of
Let
be the orthocenter of
be the circumcenter of
It is clear that is the midpoint of
is the midpoint
is the midpoint
with coefficient
Any isosceles triangle and its Gossard triangle are congruent.
Any isosceles triangle and its Gossard triangle have the same Euler line.
The Gossard triangle of the isosceles is the reflection of
in the Gossard perspector.
Denote
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Euler line of the triangle formed by the Euler line and the sides of a given triangle
Let the Euler line of meet the lines
and
at
and
respectively.
Euler line of the is parallel to
Similarly, Euler line of the
is parallel to
Euler line of the
is parallel to
Proof
Denote smaller angles between the Euler line and lines
and
as
and
respectively. WLOG,
It is known that
Let be circumcenter of
be Euler line of
(line).
Similarly,
Suppose,
which means
and
In this case
Similarly one can prove the claim in the other cases.
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Gossard triangle for triangle with angle 60
Let of the triangle
be
Let the Euler line of
meet the lines
and
at points
and
respectively.
Prove that
is an equilateral triangle.
Proof
Denote
It is known that
Therefore
is equilateral triangle.
Let be the triangle formed by the Euler lines of the
and the line
contains centroid
of the
and parallel to
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler line of the
and
the vertex
being the intersection of the Euler lines of the
and
We call the triangle as the Gossard triangle of the
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